Orbiter-Forum [IMFD] Drawing the Vector Diagram for DV Tool
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 04-24-2012, 09:57 AM #1 boogabooga Bug Crusher Drawing the Vector Diagram for DV Tool Hello, I'm trying to draw the vector diagram for a plane change + orbital circularization maneuver to get to GEO. I want to try to calculate the components by hand. This is the setup in IMFD using the DV tool: I want to plane change (RIn) 24 degrees. My apogee arrival (ApV) velocity is 1605 m/s. My required circularization velocity (Cir) is 1477. When I try to set up the vectors, I get this : (green is current orbit, red is burn vector, blue is target orbit, black is what I think the components are) When I calculate the plane change component (dVp), as shown, using the law of cosines, I get about -1282 m/s. However, as seen in the first post, the actual plane change component magnitude is much less, in the neighborhood of -950 m/s. I'm wondering if this 25% discrepancy has to do with the "vel. frame" (how the frame is being considered). I didn't completely understand this in the manual, or in the FULL manual, which I have also read. Any input would be appreciated.
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 04-25-2012, 02:43 AM #3 boogabooga Bug Crusher Thank you for your response, however, performing the maneuver is not the issue. First of all, I'm sure I want to do a combined burn: http://www.braeunig.us/space/orbmech.htm#maneuver (see section orbital plane changes) Very best would be a bi-elliptic transfer from super-synchronous orbit, but I don't want to do that. I intentionally tweaked the burn start time a little bit to improve the amount of plane change and to get ApA closer to 39786km. From here, it takes a less than 8 m/s Hohmann transfer to get to a pretty good GEO, so I don't think these errors are too significant. My main interests are calculating the vectors by hand and drawing the diagram. To calculate the segment I have labeled "dVp", I use the law of cosines, which simplifies to: 2 x (ApV + Cir) x SIN(RIn/2) = 2x(1605+1477)xSIN(24/2) = 1282 m/s This isn't even close to what you have to enter into IMFD as the "dVp" component (-948 ). The discrepancy on the order of hundreds of m/s, the best I can tell, can't be caused simply be due to starting the burn a few minutes early, etc. Actually, when I perform the burn as shown, I end up with a final velocity of 3082 m/s, which is exactly 1477+1605. So what I have labeled as "dVp" must be 1282 m/s. RIn is about 0.04, which only needs about 2 m/s to get to as close to 0 as I can get. So, the issue seems to be that this is not what IMFD considers to be dVp (Unless my math is wrong). What vector do I draw to represent dVp in IMFD? I have a feeling that this is where Vel. Frame is coming into play. You are rotating during the burn. So is the "plane change" direction constantly being redefined during the maneuver? Also, the vector that I have labeled "dVp" is not perpendicular to dVf.
 04-25-2012, 03:58 AM #4 Tommy Orbinaut What was the starting orbit? If it was circular, then the timing of the burn won't have anything to do with the ApA - that will be determined by the length of the burn. "Adjusting" the TEj of the transfer burn was a mistake - for this to work you need to center both the transfer burn and the circ burn on the nodes. That is true of ALL off-plane transfers. If you look, your current ApA is 35.58k, and the ApA after the burn would be 35.77. This difference is due to the burn NOT being properly centered on the Apo. As to the dVp vector, IMFD calculates all burns as point impulse burns (ie, dV is instant) so the vector will be at a right angle to the prograde vector at TEj=0. Therefore, there is some small inaccuracy (since the burn takes time), but the vectors don't "rotate" during the burn. I did notice that your vectors weren't at right angles - but was unsure that was intentional. dVf and dVp MUST be at right angles. Perhaps this is where the error is. By having the angle between dVf and dVp other than 90 degrees, you end up with an incorrect "burn angle", which will throw off all calculations and the "red line" will be wrong. In fact, if I had the time, I'd trig this out using basic "right triangle" trigonometry - not the law of cosines (which is only needed for non-right triangles). dVf will be one leg, the "side adjacent", the dVp will be the "side opposite" and the burn vector will be the hypotenuse. The pythagorean theorem can be used to determine the length of the hypotenuse, then dVp / dVf = tan(burnangle). Or dVp / burnvector = sine(burnangle), however you want to do it. I'm not sure it matters, but your diagram seems to be using "equatorial frame" rather than "velocity frame" - but I don't think that makes any real difference here. You'll need someone with much better math-fu to help with that. I'm thinking that this diagram is a bit more accurate - the vertical axis is the equator. The angle between the green line and the purple line is the RInc, and the burn angle = RInc * 2. I could be wrong. untitled.jpg also, ignore the typo - dFv in the diagram should be dVf. Last edited by Tommy; 04-25-2012 at 04:16 AM.

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